All right, class. So, in this video, we're going to be talking about finding proportions from raw scores. The first thing we're going to do is calculate our zcore. Next, we're going to plot our zcore on our normal curve, like we've been doing. Then, we're going to determine if what we're talking about for our particular problem is the body, tail, or middle of the distribution. We're going to look up our zcore in the unitn normal table using our column based on the body tail or middle decision that we've made. And then finally we'll multiply by the number of scores. This is an optional step and you'll see the cases in which we want to do this. So for our example question here, our data set that we have has an average income of $50,000. The standard deviation is $20,000 and the population is 1 million people. So our question is how many people make more than $100,000? So written out here we have probability of X is greater than 100,000. So our step one is going to be to calculate the Zcore. Now I've moved over here to my document where I can write things out so we can work on this together. So if we remember our first step on the slide was to calculate our Zcore. So, we remember that our zcore is going to be our x value that we're looking up minus the mean over our standard deviation. So what that gives us here is our x is 100,000, our mean is 50,000, and our standard deviation is 20,000. So, when we do all of the math here, we're going to get 50,000 over 20,000. And that's just going to equal 2.5. So, that would mean our zcore is 2.5. And we're going to remember that that is a positive 2.5. So that was step one. So now we're moving on here to step two. We're going to plot our value here on our little graph here. So where that would take us is right about here. So this is where our 2.5 would be on our zcore table here. So that's our 2.5. All right. So we are looking for things that are greater than 100,000. So that means we're going to be looking this way. So greater than 100,000. So what that means is that this is the tail portion of this graph here. So now step three we are going to go ahead and um so we determine that that is our tail and four that is going to be our looking it up on our unitn normal table. So, we're going to go in here to our unit normal table, and we're going to scroll all the way down to where we have a 2.5. So, this is a pretty big table. So, down to 2.5. So, here we go. And we decided that this was our tail value. So, this is going to be our value 0.62. Great. So let's head back to this. So that is going to be 0.06 2. Now we have that number here and that is our proportion. Yeah. So that is our proportion. So now we actually want the number of people. And so to get the number of people, we're going to multiply by the um n that we had for our sample. So we're going to multiply our number here by 1 million. And I'm going to um go ahead back here to our slides for a second and we're going to see that when we make that multiplication, this is about um 6,200 people. So that's how you would get to the final answer of our problem here. All right. So we're going to go through a second example here. So let's say we have a fictional exam. The mean of the exam was 75 out of 100 points, 75%. The standard deviation was 10 points and we'll say 100 people took this exam. So we're going to ask how many people scored higher than 90% on the exam. So, a little reminder of our steps. We're going to find the zcore, plot the score on our normal curve, determine if we're looking at a body, tail, or middle based on our question. We're going to look it up in the unitn normal table, and then we're going to multiply our proportion by the number of scores so we can get the number of people again. So, we're going to switch over here to our paper. So we are going to first calculate our zcore. So we remember that our zcore equation is going to be z= x minus the mean over the standard deviation. And that means we're looking at uh 90% on the exam. So we're going to have 90 minus our mean of 75. And that is going to be over our standard deviation of 10 right here. So that's going to be 15 over 10, which is going to be 1.5 and that is positive. So then that was our step one. So our step two is we're going to plot it on our graph here. So we're going to have our 1.5 is right about here. And that is step two. So we're going to need to go to step three. We're looking at people that scored higher than 90. So we know it's this direction here. And that means we're looking at a tail value. So then our step four is we're going to go look up our proportion on our unitn normal table. So, we're going to switch back over here to our unitn normal table and we're going to look for 1.5 and it's in the tail. So, that's going to be this number 0.0668. So, we're going to switch back over 0.0668. And then part five. Whoops. We're going to multiply this proportion. So our 0.0668 by our uh number of people in our population, which if we scroll back up here, we see that our n is 100. So we're going to multiply that times 100. And that gives us about 6. 68 people. So, sixish people. So, we're going to flip back here to our slides once again. And we'll see all of our steps once again. And that means we had again here 6.68 people. So, perfect. So then we're going to move be moving on here to two raw scores. So this is going to be a little different because we're looking at a um proportion of things between two different raw scores. So for this question, the average score on a test is a 75 with a standard deviation of five. What is the proportion of students who scored between an 83 and a 65? So in terms of our how we writing it out, we would say the proportion between so 65 and x between up to 83. So this is what we're looking at. So we're looking at somewhere between our 65 and our 83. So next what we're going to do is we're going to switch over and we're going to calculate our zcores. So now we've switched over to our paper here. So the first thing we're going to do is we're going to find the zcore for both of these numbers. So, we're going to start by finding uh 83. So, for that one, we're going to do our equation, same as we've been doing. And that's going to be our 83 minus our mean over our standard deviation. And that when you put it into your calculator is going to give you a 1.6. So then we're going to use our second zcore here and that's going to be our same equation. So then we're going to do this for 65. So we're going to do 65 minus 75 over our standard deviation of five. And you'll notice here that this is going to be a um 10 over 5. So this is going to actually be our first time that we've got a negative here. So this is positive 1.6 and this is -2. So then we're going to do the same thing that we did before. We're going to scroll down and we're going to plot our two values because we're looking for the numbers in between. So, we have 1.6 is right about here and -2 is right here. So, we're going to have this and this. So, what we're looking for is the area right here in the middle because we want between those two scores. So, we're looking at this area right here. So, the first thing we're going to do is we're going to go over to our table and we're going to find our 1.6. So, our 1.6, our positive 1.6, six. And we're looking at our middle score because we're looking at the middle part right here. We're looking at this middle part. So that's going to be giving you this right here. So that's what the middle part would be. So when we take a look back at our table, we get this 0.445. 4452. So we're going to go back here and we're going to have 0.44 52. So that's our 1.6 Zcore. And then we're going to go back over to our table once again and we're going to be looking at -2. And remember, when you're looking at this table, this is not really going to tell you the positives and negatives. So when you're looking up a zcore, this is going to be for a positive or negative one. Um, so we have our two here, and we're looking between our mean and z. So that's going to be our 0.4472 or 4772. So we're going to move back here and we have 0 4772. So that was our one for um -2. So what that is giving us is this area here. That's why we picked middle. So that's this area between Z and -2. So, what we're going to do then is because it's both of these areas together, all of this that we're interested in, we're just going to add those together. So, that'll give us a 0 9224 when we type that into our calculator. So, very good. So, this is going to be our proportion of everything that is between -2 and 1.6. So, one more thing I wanted to show you about this problem. So, if you take a look here, we've got our same two zcores. So, we're looking between -2 and 1.6. So, what we did before is we picked the middle value for each one and we were able to get this center area here. But we could have done that in a different way. So, let me show you. So what we could have done is found this right here. So this would be the um body right here and that would have given us all of this area here including this. So that would have given us this whole area. So then if we would go to our table over here and we look for 1.6 and we look at the body we get 0.9452. So if we go back and we have um our 0 94 52 area and then we're going to go over here and we see this area right here. Well, we don't want that area. We want this area in between. Right? So, if we look at this area right here at the tail part, we're looking at this area right here. So, we're going to go to our table once again and look at two and look at our tail amount. And we have 0.0228. Going to go back to here. and we're going to get 0 22 28. So, sorry, let me do that. 0.0228. And since we don't want this area, we would subtract these two numbers. And guess what? When we do that, we're going to get the same that we got up here above. So, we type this into our calculator and this is also going to be 0.9224. So, look at that. You can get the same thing using a different method here. So, we're going to do another sample question here. We have a population mean of 58 and a standard deviation of 10. So, we're going to be looking at the proportion of scores that fall between 55 and 65. So, we're going to continue on with our usual process here. We're going to find our zcores for each of these. So, that's going to be z= x minus the mean over standard deviation. So we have our first x which is 55 minus our mean which is 58 and that's going to be over our standard deviation of 10. We're going to get a -3 over 10. So that's going to give us a 0.3. So then we're going to do the same thing here. We're going to have our next score, which is 65, minus our mean of 58 over 10. And that's going to give us a 7 over 10. And that's going to give us a positive 0.7. So, there's again two ways that we can do this here. um we're going to look at our um0.3. So that's about somewhere in here vaguely. So we're going to have that. And we're going to have our 0.7, which is probably pretty close to here. And we want to find this area in between. So if you remember, we can do that in two ways. We can use the middle. And what we're going to do if we're using the middle is we're going to be figuring out so this side here. And then we're going to add this side here. together to get the full middle area. So, let's start by doing it that way. So, we're going to look for 0.3 over on our table. So, switch on over to our table and we're going to find our 0.3. There it is. And we're looking at our middle. So we have 0.1179. So we're going to go over here. 0.1179. And then we're going to do our 0.7. So back over to our table. 0.7. And we're looking at our middle once again. So that is going to be uh 0.258. Yes. So we'll go over to here and that one is 0 25 8 0. We'll add those together and that's going to give us 3759. Wonderful. Now we'll look at the other way that we can do this. We can do a body minus a tail method here. So we can look up the body. Sorry there. Had to pause for a moment. So, we're going to look up our body for our uh 0.7, which is going to be all of this. All of that. And we're going to go over to our unit normal table and we're going to look up our 0.7. So for our body, our body row for 0.7, we have 0.758. So we're going to go back over here and we have 0.7580. And then what we're going to do is we're going to take this portion here, this, and we're going to get rid of that cuz we don't want that area. We only want this middle area. That's all we want. So, we're going to go look at the tail for this one. So, we're going to go find our tail for 0.3. So over here going to find our 0.3 and our tail column is here. So that's going to give us a 0 uh 3821 and we have a 0.3 821. And remember we don't want this area. We're taking it away. We're only wanting this area. So, we're subtracting this part out. So, the number that we're going to end up getting there is going to be exactly the same as before. So, if you type that into your calculator, you're going to get 0.3759. Awesome. So, those are the two ways that you can get the exact same number here. So, then we're going to move on to another practice problem here. So for this problem, we have a population mean of 24.3, a standard deviation of 10, and this time we're asking what two scores make up the boundary for the middle 90% of the distribution. So this one's a little more difficult, but we're going to start over on our sheet here and probably hopefully make things a little bit more clear. All right, so this one we got to really reason out here. So we want the middle 90%. So if we think about that, that means we have on each side of our distribution about 5%. Right? Because if you cut off the bottom five and the top five, the middle is going to be 90%. Right? So what we're saying here is we want to find our kind of cut offs here and here we'll say because we don't know where that is yet. That's what we're solving for. So we're trying to figure out what number is this and it's going to be five% on each side. So 5% on each side. Well, we know what that means. That means our proportion on each side is going to be 0.05. Right? So that's going to be the number that we're going to find the zcore for. So we're going to be looking for this on our table. So we're going to pop back over here. So we're going to come over here to our table. And remember this time we're starting with proportion. So we're looking for the zcore. We have the proportion. So we're looking over here at the ends. So these are the tail values. So this would be the tail, the tails. So that means we're going to be looking in this tail column for a 0.05. So we're going to keep kind of scrolling down here and we're going to keep scrolling for our tail. All right, we're getting close. So, we want a 0.05 for our tail. So, the closest that we can get here is probably probably somewhere like this. So, this is 0.495. You could also say 1 uh 64, but we're going to use 1.65 because this is this is pretty close to our 0.05. 05. So either of these values probably going to work for you. So we're going to say 1.65 is our zcore. So we'll switch back over here. And we know that our z now is 1.65. So if we go back over here, this is going to be a negative 1.65 and this is a positive 1.65. 65. Cool. So, we found our zcore. So, that's all well and good. We've got our zcores, but now we have to convert those back to actual raw scores. So, if we want to do that, if you remember our equation that we had before is our zcore is equal to x minus the mean over the standard deviation. Well, we can rearrange that equation and I've done that um previously. So, if you want some assistance on how you do that, your equation rearranged with algebra is going to be mean plus Z times the standard deviation. So, that is what we rearrange our equation to be. So, we're going to solve for the first X. So this is the one that is forgative -1.65. So for that one we have our mean which we are given up here is 24.3 and then we are going to add our 1.65 65 multiplied by our standard deviation which was 10. So that was 10. So then that one is going to be 24.3 plus a 1 or 16.5. So that's just subtraction there. So what that is going to give you when you type that into your calculator is a 7.8. So then we want to solve for our other bound. So our 1.65. So our x for that is going to be same thing here plus this is going to be our positive 1.65 65 times our 10, which is our standard deviation once again. And what this is going to be is we have our same numbers here. So this is going to be our um 1.65 * 10, but it's positive this time because we're doing the positive version. So that's going to be adding 16.5 to this number. And that means our other x is going to be 40.8. eight. Awesome. So, what that means is if we go back up here to our table, this number, this raw score is 7.8 and this raw score is 40.8 and this is our mean 24.3. Right? So, that makes a lot of sense here. Very good. So now we're going to talk about what this all means. So why are we doing this? Why would you actually use this? So what you're going to do in real life is you're going to be comparing your population to your sample. So you have a population with a particular mean, a particular standard deviation. you're going to take a sample of those people and then you're going to do something like a treatment. So you'll have a control group where they won't get a treatment and you'll have the um treatment group that will get a treatment and they will have um different outcomes. So what we want to discover is is there a meaningful difference between our treated sample and our control sample. So the way that we do that is we use um exactly what we've been doing this whole time. So if you take a look here, the middle 95% here. So probability values that are near our mean of 400 indicate that our treatment has no effect. So what this would say is our sample population, our control population, so the control part of our sample, we expect them to all be in this middle 95%. Right? Because that would make no that would make sense. All of them would be pretty much average. there's nothing unique or interesting going on there. Now, when we're looking at um the treated group, we want to see that there's a big difference, a meaningful difference. So, what we're going to be looking at is the extreme 5% of each side. So, this is exactly the problem that we just did. We're looking at the extreme ends. So the 5% higher and the 5% lower at the very tip ends of this mean. So scores that are very unlikely to be attained from the original population are evidence that our treatment has an effect. So what this is saying is if you see in your treatment population values that are within those high extremes, so you see differences that are that many standard deviations above the mean, almost two standard deviations above the mean, that means we're seeing some effect here. That means it's not very likely that this treatment did not have an effect. So if you're getting values for your treatment population that is out in those farther bounds, that means your treatment has done something. So next we're going to head into a learning check here. So membership into Mensah requires a score of 130 on the Stanford Benet IQ test which has a mean of 100 and a standard deviation of 15. So what is the proportion of the population that qualifies for Mensah? So we're going to go through this one more time. So we are looking at a score of 130. We are looking at a mean of 100 and a standard deviation of 15. So what is the population that qualifies for MESA? So we take a look at this here and we're going to start with our exact same equation. Sorry about that. It's going to be Z equ= X minus the mean over our standard deviation. So that means our Z is going to be 130 minus 100 over 15 and that would be 30 over 15. And so that means our zcore is just going to be two. So we're going to go to our little graph here and that's right here at two. And we're looking for people that have 130 or above to get into Mensa. So that's going to be up here. And that is the tail portion once again. So we're going to be looking for a zcore of two. And we're looking for the tail. So, we're going to go over here onto our unit normal table. We're going to look at two in the tail. So, that's 0.0228. So, going back here, 0.0228. Awesome. So, that's our number. So, let's go back to our slides. And wonderful. That is what we have here. So then we're going to move on. True or false? It's possible to find the xcore corresponding to a percentile rank in a normal distribution. Well, we know that is true. That's kind of what we've been doing. We know that our proportion is the same as a percentile rank. So 02 would be our 20th percentile rank. Our next question, if you know a zcore, you can find the probability of obtaining that zcore in a distribution of any shape. Now, that is false. You could only do this for normal distributions. If you have some sort of funnyl looking distribution, your area under your curve is going to be different than a normal distribution. So if you have some sort of square formation or some other funky kind of um polomial that's going up and down all over the place, well, you're not going to be able to use these skills anymore from our table. You're going to have to be doing some calculus shenanigans. All right, so here's a little bit of a review here. um try to go through these by yourself and see if you can get all the answers. All right, so I hope you have gone through and tried to answer these questions. So let's just talk through these together. So when a raw score distribution is transformed into a standard distribution, what happens to its shape? Well, we know that this will have no effect. our shape will not change. So then um we're going to look at these couple different um pens here. So you have 10 pens total. Five black, three blue, one red, one green. So what is the probability of choosing a black pen? So we have five black pens over 10 pens. Our probability would be 5 over 10. So, a red or a blue pen. So, this means you could pick a red one or a blue one. So, what you would do here is red and blue both count. So, those both count as a hit. So, you would say blue is three, red is one, so that's four total. So, that would be 4 over 10, right? So, that would be a 40% chance of getting a blue or a red pen. All right. So then our next one here, a red pen and then a green pen. So this is you're pulling a red pen, putting the red pen back in, pulling the green pen, putting the green pen back in. So our chances of getting a red pen is 1 out of 10. And then we put that back in. So we have 10 total pens left. And then we have a green pen, which is one. So that' be 1 out of 10 again. And what do we do when we're doing two separate independent events like this? Well, we multiply them together. So if we multiply 1 over 10 * 1 over 10, we're going to get 1 over 100. So that would be our answer there. So then what is an assumption of random sampling? I think it's very important to say that it needs to be independent. the probability of being selected has to remain constant and you have to use replacement. So back when we did number five here with the red pen and the green pen, we have to put the red pen back in before we can pull out the green pen because they're independent. The probability has to remain constant. So you have to have the same number of pens in the bag at every time and you have to replace things once you've picked them so they could have a chance of being selected again. So that is going to be the end of this lecture here and I'll start in on the next lecture coming up. So thanks for listening.